You are now told that the function f from (a) is strictly increasing on eachinterval (4k − 1, 4k) and strictly decreasing on each interval (4k, 4k + 1).Using this, explain why f has exactly one root in each interval [4k−1, 4k]and [4k, 4k + 1].

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.x4 + x − 4 = 0,    (1, 2)f(x) = x4 + x − 4 is on the closed interval [1, 2], f(1) = , and f(2) = .

Prove that the function f (x) = cos(πx/2) + cos(πx) has a root in eachinterval [4k − 1, 4k] and [4k, 4k + 1], for every k ∈ Z. Relevant theoremsand properties of the cosine function may be freely used, if stated clearly

Let f(x) = – x2 – 4, x < 0. What is f-1?Question 4Select one:a.–√(–x – 4)b.–√(–x – 4), x < – 4c.√(x + 4)d.–√(–x – 4), x > – 4e.None of these

Analyze the trigonometric function f(x) = (4sinx + 4cosx)^2 over the interval x(-pie,pie), stating where f is increasing, decreasing, concave up, concave down, and stating the x-coordinates of all inflection points. A) Find the interval(s) on which f(x) is increasing choose: f(x) is never increasing 1 2 3 B) Find the Interval(s) on which f(x) is decreasing choose: f(x) is never idecreasing 1 2 3 c) Find the Interval(s) on which f(x) is concave up choose: f(x) is never concave up 1 2 3 d) FInd the interval(s) on which f(x) is concave down choose: f(x) is never concave down 1 2 3 e) Find the x-coordinate(s) of any/all inflection point(s). choose: f(x) has no inflection point 1 2 3

The set of values of ‘a’ for which f(x) = ax2 + 2x(1 – a) – 4 is negative for exactly three integral values of x, is